Chapter 7.3, Problem 29E

To determine

To find : the orthogonal trajectories of the family of curves and draw members of family by using graphing device.

Answer to Problem 29E

the orthogonal trajectories are the family of $y=c{x}^2$

Explanation of Solution

Given information :

The equation is $x^2+2y^2=k^2$ .

Calculation : the curves $x^2+2y^2=k^2$ form a family of circle whose center is $\left(0,\sqrt{2}\right)$ and radius is $k$ . To find a single differential equation that is satisfied by the members of the family, differentiate $x^2+2y^2=k^2$ with respect to x .

  $\begin{array}{l}{x}^2+2y^2=k^2\\ 2x+4y\frac{dy}{dx}=0\text{ [differentialte with respect to }x]\\ x+2y\frac{dy}{dx}=0\text{ [divide by 2]}\\ \frac{dy}{dx}=-\frac{x}{2y}\text{ [subtract }x\text{ and divide by }y]\end{array}$

The differential equation does not depends on k .

So, this means the slope of the tangent line at any point $\left(x,y\right)$ on one of the circle is $\frac{dy}{dx}=-\frac{x}{2y}$ .

On an orthogonal trajectory the slope of the tangent line must be negative reciprocal of this slope.

Therefore, the orthogonal trajectories must satisfy the differential equation,

  $\frac{dy}{dx}=\frac{2y}{x}$

The differential equation is separable, so to solve it

  $\begin{array}{l}\frac{dy}{dx}=\frac{2y}{x}\\ \frac{dy}{y}=2\frac{dx}{x}\text{ [separate variables]}\\ {\displaystyle \int \frac{dy}{y}={\displaystyle \int 2\frac{dx}{x}}}\text{ [integrate both side]}\\ \ln \left|y\right|=2\ln \left|x\right|+\ln \left|c\right|\text{ [simplify integrate]}\\ \ln \left|y\right|=\ln |x^2·c|\text{ [simplify]}\\ y=c{x}^2\text{ [remove ln]}\end{array}$

Where c is an arbitrary positive constant.

Thus the orthogonal trajectories are the family of $y=c{x}^2$

For members of family substitute c by different values ,

  $\begin{array}{l}y=x^2\text{ [}c=1]\\ y=-x^2\text{ [}c=-1]\\ y=2x^2\text{ [}c=2]\\ y=-2x^2\text{ [}c=-2]\end{array}$

Using implicit plotting capability of a CAS to graph the curve $y=x^2,y=-x^2,y=2x^2,y=-2x^2$

The graph is obtained as :